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Set $u=e^{iw}$ in $\cos w=z$. The resulting quadratic equation is $u^2-2zu+1=0$, with roots $z\pm i\sqrt{1-z^2}$. Their product is one, so taking a complex logarithm never encounters zero. Solving the exponential equation gives the displayed full inverse relation, with integer $n$. The principal complex logarithm and principal square root do not make both signs identical to the principal inverse cosine: at $z=0$ they yield $\pm \pi /2$, while principal $\arccos 0=\pi /2$. A single analytic branch requires compatible choices of sign and branches.

For real $\beta >1$, choose a branch of the complex logarithm whose argument interval contains $[0,2\pi /\beta ]$ strictly, and integrate $\mathrm{Log}z/(1+z^{\beta })$ over an annular sector of angle $\alpha =2\pi /\beta$. The one enclosed simple pole is $e^{i\pi /\beta }$, with residue $-i\pi e^{i\pi /\beta }/{\beta }^2$. The circular arcs tend to zero, with bounds $O(ϵ|\log ϵ|)$ and $O(R^{1-\beta }\log R)$.

Writing $I$ for the title integral and $J={\int }_0^{\infty }(1+x^{\beta }{)}^{-1}dx$, the two radial segments and the residue theorem giveDivide by $e^{i\pi /\beta }$ and equate real and imaginary parts. This gives $J=\pi /[\beta \sin (\pi /\beta )]$ and the title formula. When $1<\beta <2$, the sector extends below the negative real axis; the principal complex logarithm therefore cannot be used throughout that sector. A cut along a ray just below the positive axis avoids this issue for every $\beta >1$.

The complex number $\sqrt{3}+i$ has modulus two and principal argument $\pi /6$. Therefore its principal complex logarithm isIf the source's logarithm means the full multivalued complex logarithm, rather than its principal value, the complete answer is $\ln 2+i(\pi /6+2\pi n)$ with $n\in \mathbb{Z}$. A branch convention is necessary to obtain a unique imaginary part.

Use the principal complex logarithm at each stage of the nested complex exponentiation. Since $\mathrm{Log}i=i\pi /2$,This inner value is positive real, so its principal complex logarithm is $-\pi /2$. ConsequentlyOne must retain the parentheses: complex exponentiation is not associative. If all complex logarithm branches are allowed, the inner values are $e^{-\pi /2-2\pi k}$. Taking the outer logarithm with an independent integer branch index $\ell$ yields $\exp [-i\pi /2-2\pi ik-2\pi \ell ]=-i{e}^{-2\pi \ell }$. Thus the complete multivalued set is $\{-i{e}^{-2\pi \ell }:\ell \in \mathbb{Z}\}$, and the boxed answer selects its principal evaluation.

Put $q=|z{|}^2$ and retain the domain restriction $z\ne 0$. Then $z+1/\bar{z}=(q+1)/\bar{z}$, so the continued fraction simplifies toThe other denominators are nonzero automatically because $q+1,q+2>0$. Writing $z=x+iy$, the real-part condition becomesMultiplying by $q(q+2)$ and rearranging gives $q+4x+2=0$. Thus the first geometric locus in the complex plane is the circleIts center is $-2$ and radius is $\sqrt{2}$; it does not contain the excluded point $z=0$ on its circumference.

The principal complex logarithm has imaginary part equal to the complex argument, so its equation gives the open rayIf one interprets the logarithm as multivalued, requiring a value with imaginary part $\pi /3$ gives the same ray. Expanding $z^2$ in the other equation gives $2ixy=2ix$, hence $x(y-1)=0$. Its solution is the union of the entire imaginary axis and the horizontal line $y=1$.

These latter loci bound a right triangle with verticesThe bounded open region is $0<x<1/\sqrt{3}$, $\sqrt{3}x<y<1$. Its closure is the closed triangle. Although the logarithm is undefined at $O$, the imaginary-axis locus includes that vertex. The earlier circle is a separate locus and is not a side of this triangle.

The right angle is at $P$, so the circumcenter is the midpoint of hypotenuse $OQ$:Thus its circumcircle has the complex equationTo find the incircle, let its radius be $r$. Tangency to $x=0$ and $y=1$ puts its center at $z_i=r+i(1-r)$. The distance from that point to $y-\sqrt{3}x=0$ is $(1-r-\sqrt{3}r)/2$. Requiring this distance to be $r$ givesThereforeThis tangent circle is indeed the largest contained circle. For any contained disk of radius $\rho$, its center has distances $d_j\ge \rho$ from the three side lines. Splitting the triangle into triangles with that center as vertex gives area $\Delta =\frac{1}{2}{\sum }_j{\ell }_j{d}_j\ge s\rho$, where $s$ is its semiperimeter. Here $\Delta =1/(2\sqrt{3})$ and $s=(1+\sqrt{3})/2$, so $\rho \le \Delta /s=r$. Our circle attains this upper bound, proving maximality rather than just tangency.

Write $K={\sigma }_A(x)$ and equip $\mathcal{O}(U)$ with the compact-open topology. All algebra homomorphisms here are complex linear. Choose a finite positively oriented boundary cycle $\Gamma$ in $U\setminus K$ with winding number one on a neighborhood of $K$ and zero outside $U$. Such a cycle can be obtained as the boundary of a finite union of sufficiently small squares whose interior contains $K$ and whose closure is contained in $U$; holes are included with negative boundary orientation. Define the holomorphic functional calculus by the norm-convergent contour integralThe resolvent identity proves that $R$ is a Banach-space-valued holomorphic function off $K$. The Cauchy integral theorem, applied after each bounded linear functional, makes the integral independent of the chosen boundary cycle. The Hahn-Banach theorem then gives equality of the algebra-valued integrals. Moreover,which proves complex linearity and continuity for the compact-open topology.

For the constant function $1$ and the coordinate function $Z$, the contour can be deformed to a large circle because their integrands are holomorphic everywhere outside $K$. There the Neumann series gives $R(z)={\sum }_{n\ge 0}{x}^n{z}^{-n-1}$. Integrating term by term showsTo prove multiplicativity, take nested cycles ${\Gamma }_o,{\Gamma }_i$ around $K$, with the outer cycle winding once on the inner cycle and the inner winding zero on the outer cycle. Use the outer cycle for $f$ and the inner for $g$. The resolvent identity readsIn the resulting double integral, the term containing $R(w)$ gives $f(w)g(w)R(w)$ after the $z$ integral, by the Cauchy integral formula. The term containing $R(z)$ vanishes after the $w$ integral, because $w\mapsto g(w)/(z-w)$ is holomorphic on and inside the inner cycle's region. ThereforeThe integrals are over finite compact curves and have continuous integrands, so exchanging their order is legitimate. This proves the existence of the required continuous unital algebra homomorphism.

For uniqueness, the Runge approximation theorem gives density in $\mathcal{O}(U)$, for the compact-open topology, of rational functions with poles outside $U$. More explicitly, exhaust $U$ by compact sets that are holomorphically convex relative to $U$, and approximate on those sets by rational functions whose poles lie in the complement of $U$; this gives convergence on every compact subset. Any unital algebra homomorphism taking $Z$ to $x$ must take $(Z-\alpha {)}^{-1}$, for $\alpha \notin U$, to $(x-\alpha 1{)}^{-1}$, since it preserves inverse equations. Partial fractions and polynomial algebra therefore fix its value on every such rational function. Continuity fixes its value on their limits. Thus the continuous unital homomorphism with the prescribed coordinate value is unique. Restricting a function to a smaller neighborhood of $K$ gives the same value, either by the contour definition or this uniqueness argument.

To prove the holomorphic spectral mapping theorem, first let $\mu \notin f(K)$. There is an open neighborhood $V\subseteq U$ of $K$ on which $f-\mu$ has no zeros. The holomorphic function $h=1/(f-\mu )$ on $V$ supplies a two-sided inverse throughThus ${\sigma }_A({\Theta }_x(f))\subseteq f(K)$. Conversely, if $\lambda \in K$ and $\mu =f(\lambda )$, the divided differenceis holomorphic on $U$, and multiplicativity givesThe two factors commute. A commuting product $uv$ can be invertible only if both factors are invertible: its inverse commutes with each factor, and $v(uv{)}^{-1}$ is a two-sided inverse for $u$. Invertibility of the displayed product would therefore contradict $\lambda \in K$. This proves

For the root, let $D=\mathbb{C}\setminus (-\infty ,0]$ and $q(z)=\exp (\mathrm{Log}z/3)$, using the principal complex logarithm. Define $y={\Theta }_x(q)$. Since $q(z{)}^3=z$, multiplicativity and the holomorphic spectral mapping theorem giveThis is the principal cube root of a Banach-algebra element.

For uniqueness, suppose $v^3=x$ and ${\sigma }_A(v)\subset S$. Cubing maps the whole open sector $S$ into $D$, sois a continuous unital algebra homomorphism: pullback by cubing is continuous for the compact-open topology, since images of compact sets are compact. It takes the coordinate function to $v^3=x$, and uniqueness already proved gives $F={\Theta }_x$. For $z\in S$, the scalar principal cube root satisfies $q(z^3)=z$, because $3\arg z\in (-\pi ,\pi )$. HenceThis proves uniqueness with the stated spectral sector, without assuming the ambient Banach algebra is commutative or that algebra characters separate its elements.

For $|w|\le 1/2$, the power series of the principal complex logarithm is absolutely convergent and givesThus $C_1=2$ works. The value $1-w$ stays in the right half-plane, so there is no logarithm branch ambiguity. This is the bounds for the principal logarithm near one estimate used to control product tails.

The printed assertion is false. Take real $w=1-\epsilon$ with $0<\epsilon <1$. The principal complex logarithm is unambiguous, butNo finite $C_3$ can satisfy the proposed estimate, even on these points within the logarithm's domain. At $w=1$ the logarithm is undefined, and the usual principal analytic branch also excludes its cut. These are additional reasons the literal all-$w$ assertion cannot hold.

The valid estimate for the final product proof is insteador equivalently $|1-w|\le e^{|w|}$ for every $w$, including $w=1$. The upper bound on the real part of the logarithm is enough for growth; a bound on the full modulus of the logarithm is unnecessary. The preceding root Solution supplies the complete convergence and exponential-growth proof using this correction.

For the unrestricted zero sequence, introduce the Weierstrass elementary factorsFor $|u|\le 1/2$, cancellation of the first $p$ terms of the logarithm power series gives a holomorphic logarithm of the factor:Define the canonical productFix any compact disc $|z|\le R$. Since $|z_n|\to \infty$, eventually $|z/z_n|\le 1/2$ uniformly on this disc. The tail logarithms are then bounded by $2(1/2{)}^{n+1}$, a summable bound independent of $z$. Their sum is holomorphic by locally uniform convergence of holomorphic functions, and its exponential is nowhere zero. Multiplication by the finite initial product therefore gives an entire function. Each factor has a simple zero exactly at its specified $z_n$, the exponential factor has none, and a tail beginning beyond any given compact disc is nonvanishing there. Thus $f$ has precisely the prescribed zeros and no others, each simple, and $f(0)=1$. This proves the required existence without an assumption on ${\sum }_n|z_n{|}^{-1}$.

For the final, unheaded product request, suppose now that ${\sum }_n|z_n{|}^{-1}<\infty$. On $|z|\le R$, all sufficiently late factors have $|z/z_n|\le 1/2$, so part (a) bounds their principal complex logarithms by $2R/|z_n|$. The logarithms converge absolutely and uniformly; infinite product convergence from logarithmic tails proves thatis entire and again has exactly the prescribed simple zeros. For its finite partial products,Passing to the limit provesIn particular $P$ is an entire function of exponential type. This argument uses a valid modulus estimate and does not require the false logarithm assertion in part (c).

A branch of the complex logarithm must be specified to obtain a unique locus. First use the principal complex logarithm. Away from the negative real axis, write $z=r{e}^{i\theta }$ with $r>0$ and $-\pi <\theta <\pi$. Then the complex logarithms of $z$ and its complex conjugate are $\log r+i\theta$ and $\log r-i\theta$. Equating real and imaginary parts gives $\log r=\theta$. Consequently the principal-branch locus is the open segment of a logarithmic spiralThe origin is excluded because its complex logarithm does not exist. If the principal value is assigned argument $\pi$ on the negative real axis, a point there equals its own complex conjugate, so its principal complex logarithm would have to equal $i$ times itself. That forces the complex logarithm to be zero and cannot hold there; the endpoints remain excluded.

If the printed equation instead means that some values of the multivalued complex logarithms satisfy it, write those values as $\log r+it$ and $\log r+iu$, with $e^{it}=z/r$ and $e^{iu}=\overline{z}/r$. Equality gives $\log r=-u=t$, hence $u=-t$, which is compatible with conjugation. Every real $t$ is allowed. Under this interpretation the locus is the entire logarithmic spiral $z=e^{(1+i)t}$, $t\in \mathbb{R}$. The PDF does not choose between these branch conventions.

For a nonzero complex number, write $z=r{e}^{i\theta }$ with $r=|z|>0$ and principal complex argument $-\pi <\theta \le \pi$. The principal complex logarithm and corresponding complex exponentiation areThe value at zero is not defined by this formula. The choice of complex argument matters: these are single-valued principal powers, rather than all values of a multivalued complex logarithm.

For the requested sketch, $(1+i)\mathrm{Log}z=(\log r-\theta )+i(\log r+\theta )$. Its complex exponential has complex modulus $e^{\log r-\theta }$, soThis is one turn of a logarithmic spiral, with increasing radius as the polar angle increases. It passes through $1$ at $\theta =0$, approaches the excluded endpoint $-e^{-\pi }$ from below the negative real axis, and ends at the included endpoint $-e^{\pi }$ from above that axis. The endpoint distinction follows from the chosen principal complex argument.

Let $w$ denote the value of the complex logarithm used in the equation. Factoring $w^2+{\pi }^2/4$ gives $w=i\pi /2$ or $w=-i\pi /2$. Since $z=e^w$, the complete answer isThe principal complex logarithm takes exactly these values at $i$ and $-i$, so both are solutions with that convention. If the complex logarithm is interpreted as multivalued, the same answer holds in the sense that a logarithm value satisfies the equation. The equation is not an assertion that every logarithm value of these numbers has the stated square.

On the stated branch of the principal complex logarithm, the only denominator zero is $z=1$, and $\log z=(z-1)+O((z-1{)}^2)$. ThereforeZero is a logarithmic branch point, not a removable isolated singularity, even though $1/\log z$ tends to zero as $z\to 0$ within the cut plane. Continuation around zero changes $\log z$ by $2\pi i$. The nonpositive real axis is the chosen branch cut; its nonzero points are not intrinsically branch points and admit local analytic continuation. There are no other finite singularities inside the specified branch domain.

It helps to separate the tangent and the squaring map. Put $t=\tan (z/2)$, with $z=x+iy$. The identityshows that $0<x<\pi /2$ gives $\mathrm{Re}t>0$ and $|t|<1$. To see that every point of this right half-disc occurs, setThen $z=-i\mathrm{Log}R$ has $0<\mathrm{Re}z<\pi /2$ and satisfies $\tan (z/2)=t$. The principal complex logarithm is unambiguous because $R$ is in the first quadrant.

Squaring maps this right half-disc bijectively onto the unit disc with the interval $(-1,0]$ removed: the argument of $t$ lies between $-\pi /2$ and $\pi /2$, and doubles into $(-\pi ,\pi )$. Thus the desired image isNeither derivative vanishes in the domain, so the composition is also a conformal map onto that slit disc.

The residue theorem states that a meromorphic function with finitely many poles inside a positively oriented simple closed contour and none on it satisfies ${\oint }_{C}F(z)dz=2\pi i\sum \mathrm{Res}(F;a)$. The function must be holomorphic on a neighbourhood of the contour and its interior away from those poles.

Write $J={\int }_0^{\infty }{x}^{1/2}\log x/(1+x^2)dx$ and $I={\int }_0^{\infty }{x}^{1/2}/(1+x^2)dx$. Both integrals converge absolutely. Use the principal complex logarithm in the upper half-plane, with $z^{1/2}=e^{\log z/2}$, and an upper semicircle of radius $R>1$ indented above zero by a clockwise semicircle of radius $ϵ<1$. The branch values on the negative real side are upper limits; equivalently use contours arbitrarily slightly above that side before taking a limit.

For $F(z)=z^{1/2}\log z/(1+z^2)$, the large arc is $O(R^{-1/2}\log R)$ and the small arc is $O({ϵ}^{3/2}|\log ϵ|)$, so both vanish. On $z=-x$ approached from above, $z^{1/2}=i\sqrt{x}$ and $\log z=\log x+i\pi$. The oriented negative segment therefore contributes $iJ-\pi I$, and the positive segment contributes $J$. The only enclosed pole is $i$, with residueConsequentlyComparing imaginary and real parts yields the upper-half-plane contour for square-root logarithmic integrals evaluation:

For $z\ne 0$, the principal value of complex exponentiation is $z^w=\exp (w\mathrm{Log}z)$, where the principal complex logarithm is $\mathrm{Log}z=\log |z|+i\mathrm{Arg}z$ with $-\pi <\mathrm{Arg}z\le \pi$. De Moivre's theorem states that $(\cos \theta +i\sin \theta {)}^n=\cos (n\theta )+i\sin (n\theta )$ for every integer $n$.

Since $\sqrt{3}+i=2e^{i\pi /6}$, the six roots are

Use the principal complex logarithm, for which $\mathrm{Log}i=i\pi /2$, and setThen the definition of complex exponentiation givesand thereforeOther choices of branch of the complex logarithm produce further valid answers.

Write $z=r{e}^{i\theta }$ with $r>0$. On a fixed branch of the complex logarithm on which $\log z=\log r+i\theta$ and $\log \bar{z}=\log r-i\theta$, the equation becomesIts real and imaginary parts give the same condition $\log r=\theta$, orThis is a logarithmic spiral. For the principal complex logarithm, it is the portion parametrized by $-\pi <\theta <\pi$ that avoids the branch cut.

The number $i{e}^{i\varphi }$ has modulus one and argument $\varphi +\pi /2$ modulo $2\pi$. Therefore its complex logarithm has valuesEvery value has real part zero. For the principal complex logarithm, choose the unique integer giving imaginary part in $(-\pi ,\pi ]$. The unwrapped value $\varphi +\pi /2$ is the principal imaginary part only when it already lies in that interval.

Using the principal complex logarithm, $\mathrm{Log}i=i\pi /2$ and $\mathrm{Log}(-e)=1+i\pi$. Thus the values and their real parts and imaginary parts areFor the final row, first calculate $i^i=e^{-\pi /2}$ and only then raise this positive real number to the power $i$. Treating nested powers as ordinary real powers would ignore the branching of complex exponentiation.

If all branches are requested, $\sqrt{i}$ also has the negative of its principal value, and the complex logarithm of $-e$ is $1+i(\pi +2\pi k)$. Allowing nonprincipal logarithms of the positive base $2$ gives $e^{-2\pi k}{e}^{i\ln 2}$. Finally, all values of $(i^i{)}^i$ are $-i{e}^{-2\pi m}$, for integers $k,m$ as appropriate. These alternatives do not alter the principal-value table.